Problem 25 - Algebra and Trigonometry

Question

Find all real solutions of the equation. \(x^{2}-2 x-15=0\)

Step-by-Step Solution

Answer

The real solutions are \( x = 5 \) and \( x = -3 \).

Step 1: Identify the Equation Type

The given equation is a quadratic equation of the form \[ ax^2 + bx + c = 0 \] where, in this equation, \( a = 1 \), \( b = -2 \), and \( c = -15 \).

Step 2: Use the Quadratic Formula

To find the roots of the quadratic equation, we use the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \( a = 1 \), \( b = -2 \), and \( c = -15 \) into the formula.

Step 3: Calculate the Discriminant

First, calculate the discriminant part of the formula \[ b^2 - 4ac \]Substitute the values: \[ (-2)^2 - 4(1)(-15) = 4 + 60 = 64 \].The discriminant is 64.

Step 4: Calculate the Solutions

Since the discriminant is positive, there are two real solutions:Substitute back into the formula:\[ x = \frac{-(-2) \pm \sqrt{64}}{2(1)} = \frac{2 \pm 8}{2} \].Calculate each solution:1. \( x = \frac{2 + 8}{2} = 5 \)2. \( x = \frac{2 - 8}{2} = -3 \)

Step 5: Verify the Solutions

Substitute \( x = 5 \) and \( x = -3 \) back into the original equation to verify:For \( x = 5 \):\[ 5^2 - 2(5) - 15 = 25 - 10 - 15 = 0 \]For \( x = -3 \):\[ (-3)^2 - 2(-3) - 15 = 9 + 6 - 15 = 0 \]Both values satisfy the original equation.

Key Concepts

Quadratic Formula

When faced with a quadratic equation like \(x^2 - 2x - 15 = 0\), the quadratic formula is your go-to tool. This formula is a universal method used to find the roots of any quadratic equation, which is often in the form \(ax^2 + bx + c = 0\).
In this equation, we have \(a = 1\), \(b = -2\), and \(c = -15\). The quadratic formula is expressed as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

By substituting the given values into the formula, you will begin solving for \(x\). The symbol \(\pm\) indicates that there are typically two solutions, a fundamental property of quadratic equations. Understanding this formula is crucial as it provides the foundation for finding real solutions.

Discriminant Calculation

Before extracting the roots using the quadratic formula, it is pivotal to compute the discriminant. The discriminant, represented as \(b^2 - 4ac\), is located under the square root in the quadratic formula and influences the nature of the roots.
For the equation \(x^2 - 2x - 15 = 0\), calculate the discriminant:

• \((-2)^2 - 4 \times 1 \times (-15) = 4 + 60 = 64\)

A positive discriminant, like 64, indicates that the quadratic equation has two distinct real solutions. Conversely, a zero discriminant results in a single real solution, and a negative one implies complex solutions. Thus, the discriminant tells you what to expect before even solving the equation.

Real Solutions

Once the discriminant is computed, you can determine the roots or solutions of the equation. Since our discriminant of 64 is positive, there are two real solutions for the equation \(x^2 - 2x - 15 = 0\).
Substitute back into the quadratic formula to find these solutions:

• First: \( x = \frac{2 + 8}{2} = 5 \)
• Second: \( x = \frac{2 - 8}{2} = -3 \)

These calculations provide the real values of \(x\) that satisfy the equation. Solving it step-by-step using the quadratic formula ensures a clear derivation of results from start to finish.

Verifying Solutions

Even after calculating the solutions, verification is an essential step. Verifying solutions means substituting them back into the original equation to ensure that they truly satisfy it. This practice eliminates the risk of calculation errors.
For \(x = 5\):

• \(5^2 - 2(5) - 15 = 25 - 10 - 15 = 0\)

For \(x = -3\):

• \((-3)^2 - 2(-3) - 15 = 9 + 6 - 15 = 0\)

Both solutions yield values that are true for the equation, confirming their correctness. Verifying solutions is a prudent step in mathematical problem-solving, instilling confidence in your answers.